Here are some important 1-qubit gates. You have the three Pauli matrices:
\begin{align*}
X &= \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}, & Y &= \begin{pmatrix} 0 & -i \\ i & 0 \end{pmatrix}, & Z &= \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}
\end{align*}
You also have the Hadamard gate
\begin{equation*}
H = \frac{1}{\sqrt{2}} \begin{pmatrix} 1 & 1 \\ 1 & -1 \end{pmatrix}
\end{equation*}
It is true that
\begin{equation*}
H = \frac{1}{\sqrt{2}} (X + Z)
\end{equation*}
but I am not sure how the sum of two gates can be implemented. Indeed, if you interpret as the sum of two gates as applying them in parallel, then you cannot do that with a single qubit because that would involve cloning. Besides, the factor \(1/\sqrt{2}\) is not unitary.
Two more gates are \(S\) and \(T\):
\begin{align*}
S &= \begin{pmatrix} 1 & 0 \\ 0 & i \end{pmatrix}, & T & = \begin{pmatrix} 1 & 0 \\ 0 & \exp{(i \pi / 4)} \end{pmatrix}
\end{align*}
The \(T\) gate is the square root of the \(S\) gate:
\begin{equation*}
S = T T
\end{equation*}
The \(S\) gate in turn is the square root of the \(Z\) gate:
\begin{equation*}
Z = S S = T T T T
\end{equation*}
Note that \(S\) and \(T\) are not hermitian but they are unitary.
You can write an element of a subspace of matrices spanned by all powers of \(T\):
\begin{equation*}
a I + b T + c S + d ST + e Z + f ZT + g ZS + h ZST
\end{equation*}
This is closed under addition and multiplication. Note that
\begin{align*}
T^{\dagger} &= Z S T, & S^{\dagger} &= Z S, & Z^{\dagger} &= Z, & (S T)^{\dagger} &= Z T
\end{align*}
The eigen-decomposition of \(X\) and \(Y\) involve \(H\), \(Z\), and \(S\):
\begin{align*}
X &= H Z H = H T T T T H, & Y &= S H Z H S^{\dagger} = T T H T T T T H T^{\dagger} T^{\dagger}
\end{align*}
This means that in principle you only need \(H\) and \(T\) (and \(T^{\dagger}\)) to generate the Pauli matrices.
What about the eigen-decomposition of \(H\)? Well, you get
\begin{align*}
H & = G Z G^{\dagger}, & G &= \begin{pmatrix} \cos{(\pi/8)} & -\sin{(\pi/8)} \\ \sin{(\pi/8)} & \cos{(\pi/8)} \end{pmatrix}, & \cos{(\pi/8)} &= \frac{\sqrt{2 + \sqrt{2}}}{2}, & \sin{(\pi/8)} &= \frac{\sqrt{2 - \sqrt{2}}}{2}
\end{align*}
Can you write \(G\) in terms of \(T\)? Note that both \(S\) and \(T\) can be re-written in a similar way:
\begin{align*}
S &= \exp{(i \pi / 4)} \begin{pmatrix} \exp{(-i \pi /4)} & 0 \\ 0 & \exp{(i \pi / 4)} \end{pmatrix}, & T &= \exp{(i \pi / 8)} \begin{pmatrix} \exp{(-i \pi /8)} & 0 \\ 0 & \exp{(i \pi / 8)} \end{pmatrix}
\end{align*}
Indeed, \(Z\) can also be re-written in this way:
\begin{equation*}
Z = \exp{(i \pi / 2)} \begin{pmatrix} \exp{(-i \pi /2)} & 0 \\ 0 & \exp{(i \pi / 2)} \end{pmatrix}
\end{equation*}
When written in this way, it seems plausible that \(G\) can be written in terms of \(T\). Introduce two new unitary matrices:
\begin{align*}
P(\theta) &= \exp{(i\theta)} \begin{pmatrix}
1 & 0 \\
0 & 1
\end{pmatrix}, & Q(\theta) &= \begin{pmatrix}
\exp{(-i\theta)} & 0 \\
0 & \exp{(i\theta)}
\end{pmatrix}
\end{align*}
Then
\begin{align*}
T &= P(\pi/8) Q(\pi/8), & S &= P(\pi/4) Q(\pi/4), & Z &= P(\pi/2) Q(\pi/2)
\end{align*}
Note that
\begin{equation*}
P(\theta) + \left[ P(\theta) \right]^{\dagger} = Q(\theta) + \left[ Q(\theta) \right]^{\dagger} = 2 \cos{(\theta)} \begin{pmatrix}
1 & 0 \\
0 & 1
\end{pmatrix}
\end{equation*}
However, addition of unitary quantum gates requires parallel wires.
